SPOJ QTREE Query on a tree(树链剖分）

SPOJ QTREE Query on a tree

题目链接

树链剖分基础题

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 10005;

int dep[N], top[N], sz[N], son[N], fa[N], id[N], idx, val[N];
int n;
vector<int> g[N];

struct Edge {
	int u, v, val;
	void read() {
		scanf("%d%d%d", &u, &v, &val);
	}
} e[N];

void dfs1(int u, int f, int d) {
	dep[u] = d;
	sz[u] = 1;
	son[u] = 0;
	fa[u] = f;
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == f) continue;
		dfs1(v, u, d + 1);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v])
			son[u] = v;
	}
}

void dfs2(int u, int tp) {
	top[u] = tp;
	id[u] = ++idx;
	if (son[u]) dfs2(son[u], tp);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (v == fa[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)
const int INF = 0x3f3f3f3f;

struct Node {
	int l, r, val;
} node[N * 4];

void pushup(int x) {
	node[x].val = max(node[lson(x)].val, node[rson(x)].val);
}

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r;
	if (l == r) {
		node[x].val = val[l];
		return;
	}
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
	pushup(x);
}

void add(int v, int val, int x = 0) {
	if (node[x].l == node[x].r) {
		node[x].val = val;
		return;
	}
	int mid = (node[x].l + node[x].r) / 2;
	if (v <= mid) add(v, val, lson(x));
	if (v > mid) add(v, val, rson(x));
	pushup(x);
}

int query(int l, int r, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		return node[x].val;
	}
	int mid = (node[x].l + node[x].r) / 2;
	int ans = 0;
	if (l <= mid) ans = max(ans, query(l, r, lson(x)));
	if (r > mid) ans = max(ans, query(l, r, rson(x)));
	return ans;
}

int gao(int u, int v) {
	int tp1 = top[u], tp2 = top[v];
	int ans = 0;
	while (tp1 != tp2) {
		if (dep[tp1] < dep[tp2]) {
			swap(tp1, tp2);
			swap(u, v);
		}
		ans = max(query(id[tp1], id[u]), ans);
		u = fa[tp1];
		tp1 = top[u];
	}
	if (u == v) return ans;
	if (dep[u] > dep[v]) swap(u, v);
	ans = max(query(id[son[u]], id[v]), ans);
	return ans;
}

int T;

int main() {
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		idx = 0;
		for (int i = 1; i <= n; i++) g[i].clear();
		for (int i = 1; i < n; i++) {
			e[i].read();
			g[e[i].u].push_back(e[i].v);
			g[e[i].v].push_back(e[i].u);
		}
		dfs1(1, 0, 1);
		dfs2(1, 1);
		char Q[10];
		for (int i = 1; i < n; i++) {
			if (dep[e[i].u] < dep[e[i].v]) swap(e[i].u, e[i].v);
			val[id[e[i].u]] = e[i].val;
		}
		build(1, idx);
		int a, b;
		while (scanf("%s", Q)) {
			if (Q[0] == 'D') break;
			scanf("%d%d", &a, &b);
			if (Q[0] == 'Q') printf("%d\n", gao(a, b));
			if (Q[0] == 'C') add(id[e[a].u], b);
		}
	}
	return 0;
}